tracediagramreview.tex
\subsection{Gluing Lemmas}\label{s:gluinglemmas}

This section introduces some additional notation that will be used to generate the recurrence formulas in the later sections.

Finally, spin networks satisfy certain recoupling'' identities. These are usually written in terms of the $6j$-symbols described next, but we will introduce notation useful in simplifying the specific recurrence relations we obtain later in the paper.

\begin{definition}[$6j$-symbols, Definition 5.4 in \cite{LP}]\label{d:6jsymbol}
The \emph{$6j$-symbols} are the coefficients in the following change-of-basis equation:
$$\tikz[trivalent]{ \coordinate(vxa)at(0,.35) edge[]node[leftlabel,pos=1]{d}(0,-.15)edge[bend left]node[leftlabel,pos=1]{a}(-.6,1.15); \coordinate(vxb)at(.3,.75) edge[bend left]node[basiclabel,pos=.7,right=2pt]{e}(vxa) edge[bend left]node[leftlabel,pos=1]{b}(0,1.15)edge[bend right]node[rightlabel,pos=1]{c}(.6,1.15);} = \sum_{f\in\lceil a,b\rfloor\cap\lceil c,d\rfloor} \sixj abdcef \tikz[trivalent]{ \coordinate(vxa)at(0,.35) edge[]node[leftlabel,pos=1]{d}(0,-.15)edge[bend right]node[rightlabel,pos=1]{c}(.6,1.15); \coordinate(vxb)at(-.3,.75) edge[bend right]node[basiclabel,pos=.7,left=3pt]{f}(vxa) edge[bend right]node[rightlabel,pos=1]{b}(0,1.15)edge[bend left]node[leftlabel,pos=1]{a}(-.6,1.15);} .$$
\end{definition}
Note that this definition differs slightly from those found in the literature \cite{CFS95,Kau91}.

Together with the fusion identity, this provides an identity for gluing a strand across'' a vertex.
\begin{proposition}[Strand-Vertex Gluing]\label{l:vertexglue}
$$\tikz[trivalent]{ \coordinate(vxa)at(.5,.5) edge[]node[rightlabel,pos=1]{a}(.5,-.4) edge[bend left]node[rightlabel,pos=1]{c}(0,1.1) edge[bend right]node[rightlabel,pos=1]{b}(1,1.1); \draw[blue](.3,-.4)to(.3,-.1)to[wavyup](-.2,1.1)node[leftlabel]{d};} = \sum_{e\in\iadm{a,d}} \mff_e(a,d) \tikz[trivalent]{ \coordinate(vxa)at(0,.35) edge[blue,bend left]node[leftlabel,pos=1]{d}(-.6,1.15); \coordinate(vxc)at(0,-.15) edge[]node[leftlabel]{e}(vxa) edge[blue,bend right]node[leftlabel,pos=1]{d}(-.3,-.55)edge[bend left]node[rightlabel,pos=1]{a}(.3,-.55); \coordinate(vxb)at(.3,.75) edge[bend left]node[basiclabel,pos=.7,right=2pt]{a}(vxa) edge[bend left]node[leftlabel,pos=1]{c}(0,1.15)edge[bend right]node[rightlabel,pos=1]{b}(.6,1.15);} = \sum_{\substack{e\in\iadm{a,d}\\ f\in\iadm{c,d}\cap\iadm{b,e}}} \mff_e(a,d)\sixj dcebaf \tikz[trivalent]{ \coordinate(vxa)at(0,.35) edge[bend right]node[rightlabel,pos=1]{b}(.6,1.15); \coordinate(vxc)at(0,-.15) edge[]node[leftlabel]{e}(vxa) edge[blue,bend right]node[leftlabel,pos=1]{d}(-.3,-.55)edge[bend left]node[rightlabel,pos=1]{a}(.3,-.55); \coordinate(vxb)at(-.3,.75) edge[bend right]node[basiclabel,pos=.7,left=3pt]{f}(vxa) edge[bend right]node[rightlabel,pos=1]{c}(0,1.15)edge[blue,bend left]node[leftlabel,pos=1]{d}(-.6,1.15);} .$$
\begin{proof}
The first step applies the fusion identity (Proposition \ref{p:bubblefusion}) to the lower part of the diagram, introducing a summation over $e$. The second step uses the change-of-basis formula (Definition \ref{d:6jsymbol}) on the upper half of the diagram.
\end{proof}
\end{proposition}

We now introduce specialized notation for the case of the above lemma with $d=1$, since it will simplify the expression of the product in Theorem \ref{t:simplerecurrence}.

\begin{definition}\label{d:fusioncoeff}
Given an admissible triple $\{a,b,c\}$, $a'\in\iadm{1,a}$, and $c'\in\iadm{1,c}$, define the \emph{fusion coefficient} $\mathfrak{F}$ and the \emph{normalized fusion coefficient} $\hat{\mathfrak{F}}$ by
\begin{align*}
\Fus ba{a'}c{c'} \equiv \mfs_{a'}(1,a) \mff_{a'}(1,a) \sixjt 1c{a'}ba{c'}.\\
\nFus ba{a'}c{c'} \equiv \mfs_{a'}(1,a) \sqrt{\tfrac{\mff_{a'}(1,a)}{\mff_{c'}(1,c)}} \sixjt 1c{a'}ba{c'}.
\end{align*}
\end{definition}

It is immediate that
$$\label{eq:fusnormalization} \Fus ba{a'}c{c'}=\sqrt{\mff_{a'}(1,a)\mff_{c'}(1,c)} \nFus ba{a'}c{c'}.$$
Also, $a'\in\iadm{1,a}$ is equivalent to requiring $a'=a\pm1$, so given $\{a,b,c\}$ there are four choices for $\{a',c'\}$. Table \ref{t:fusioncoeff} shows the values of the $6j$-symbol, the fusion coefficient, and the reduced coefficient in each of the four cases. Observe that both the fusion coefficient and its normalized version are invariant under the interchange of $a$ and $c$:
$$\Fus ba{a'}c{c'}=\Fus bc{c'}a{a'} \quad\text{and}\quad \nFus ba{a'}c{c'}=\nFus bc{c'}a{a'}.$$
These values are taken from Corollary 5.5 in \cite{LP}, together with the fact that the additional fusion constant is either $\mff_{a+1}(1,a)=1$ or $\mff_{a-1}(1,a)=\frac{a}{a+1}$. This table corrects sign errors in the $6j$-symbol formulas provided on pp. 79-80 of \cite{Pet06}.

\def\spacer{\tikz{\draw[white,transparent](0,-.2)--(0,1.2);}}
\def\spacerb{\tikz{\draw[white,transparent](0,-.4)--(0,1.4);}}
\begin{center}
\begin{table}
\begin{tabular}{|c|ccc|}
\hline
\spacer
$\{a',c'\}$ & $\sixjt 1c{a'}ba{c'}$ & $\Fus ba{a'}c{c'}$ & $\nFus ba{a'}c{c'}$ \\\hline
\spacer
$\{a+1,c+1\}$ & $1$ & $1$ & $1$\\
\spacer
$\{a-1,c+1\}$ & $\tfrac{\mfe_c(a,b)}{a}$ & $-\tfrac{\mfe_c(a,b)}{a+1}$
& $-\tfrac{\mfe_c(a,b)}{\sqrt{a(a+1)}}$ \\
\spacer
$\{a+1,c-1\}$ & $-\tfrac{\mfe_a(c,b)}{c+1}$ & $-\tfrac{\mfe_a(c,b)}{c+1}$
& $-\tfrac{\mfe_a(c,b)}{\sqrt{c(c+1)}}$ \\
\spacerb
$\{a-1,c-1\}$ & $\frac{\mfe_b(a,c)(\mfe(a,b,c)+1)}{a(c+1)}$ & $-\tfrac{\mfe_b(a,c)(\mfe(a,b,c)+1)}{(a+1)(c+1)}$
& $-\tfrac{\mfe_b(a,c)(\mfe(a,b,c)+1)}{\sqrt{a(a+1)c(c+1)}}$ \\\hline
\end{tabular}
\caption{The four basic fusion coefficients, with corresponding $6j$-symbols and normalized values.}\label{t:fusioncoeff}
\end{table}
\end{center}

With a different orientation, Proposition \ref{l:vertexglue} becomes
\begin{proposition}\label{l:gluevertex-alt}
$$\label{eq:glueonesum2} \tikz[trivalent]{ \coordinate(vxa)at(.5,.4)edge[]node[rightlabel,pos=1]{b}(.5,-.1) edge[bend left]node[leftlabel,pos=1]{c}(0,1.1) edge[bend right]node[rightlabel,pos=1]{a}(1,1.1); \draw[blue](.2,1.1)to[out=-90,in=-90,looseness=2](.8,1.1)node[leftlabel]{1};} = \sum_{\substack{a'\in\iadm{1,a}\\ c'\in\iadm{1,c}\cap\iadm{b,a'}}} \Fus ba{a'}c{c'} \tikz[trivalent]{ \coordinate(vxa)at(.5,.4)edge[]node[rightlabel,pos=1]{b}(.5,-.1); \coordinate(vxb)at(.1,.8)edge[bend right]node[leftlabel]{c'\:}(vxa) edge[bend left]node[leftlabel,pos=1]{c}(-.2,1.2) edge[blue,bend right]node[leftlabel,pos=1]{1}(.4,1.2); \coordinate(vxc)at(.9,.8)edge[bend left]node[rightlabel]{a'}(vxa) edge[blue,bend left]node[rightlabel,pos=1]{1}(.6,1.2) edge[bend right]node[rightlabel,pos=1]{a}(1.2,1.2);}.$$
\begin{proof}
In terms of the fusion coefficient, Proposition \ref{l:vertexglue} is
$$\label{eq:gluevertexleft} \tikz[trivalent]{ \coordinate(vxa)at(.5,.5) edge[]node[rightlabel,pos=1]{a}(.5,-.4) edge[bend left]node[rightlabel,pos=1]{c}(0,1.1) edge[bend right]node[rightlabel,pos=1]{b}(1,1.1); \draw[blue](.3,-.4)to(.3,-.1)to[wavyup](-.2,1.1)node[leftlabel]{1};} = \sum_{\substack{a'\in\iadm{a,1}\\ c'\in\iadm{c,1}\cap\iadm{b,a'}}} \mfs_{a'}(a,1)\Fus ba{a'}c{c'} \tikz[trivalent]{ \coordinate(vxa)at(0,.35) edge[bend right]node[rightlabel,pos=1]{b}(.6,1.15); \coordinate(vxc)at(0,-.15) edge[]node[leftlabel]{a'}(vxa) edge[bend right]node[blue,leftlabel,pos=1]{1}(-.3,-.55)edge[bend left]node[rightlabel,pos=1]{a}(.3,-.55); \coordinate(vxb)at(-.3,.75) edge[bend right]node[basiclabel,pos=.7,left=3pt]{c'}(vxa) edge[bend right]node[rightlabel,pos=1]{c}(0,1.15)edge[blue,bend left]node[leftlabel,pos=1]{1}(-.6,1.15);}.$$
Reflect this relation vertically using Proposition \ref{p:spinnetreflection}, and extend the strands labeled by $1$ and $c$ on both sides of the equation to obtain:
$$\label{eq:extendstrands} \tikz[trivalent]{ \coordinate(vertex)at(.5,.5)edge[]node[rightlabel,pos=1]{a}(.5,1) edge[bend right](.1,.2)edge[bend left]node[rightlabel,pos=1]{b}(1,0); \draw[blue](-.1,.2)to[wavyup](.3,1); \draw[red](-.1,.2)to[out=-90,in=-90](-.5,.2)to[wavyup](-.4,1)node[rightlabel]{1} (.1,.2)to[out=-90,in=-90,looseness=1.5](-.7,.2)to[wavyup](-.6,1)node[leftlabel]{c}; }=\sum_{\substack{a'\in\iadm{a,1}\\ c'\in\iadm{c,1}\cap\iadm{b,a'}}} \mfs_{a'}(a,1)\Fus ba{a'}c{c'} \tikz[trivalent]{ \coordinate(vertexa)at(.2,.2)edge[blue,bend right](-.1,0)edge[bend left](.3,0); \coordinate(vertexb)at(.5,.5)edge[bend right]node[leftlabel,pos=.3]{c'}(vertexa) edge[bend left]node[rightlabel,pos=1]{b}(1.1,-.2); \coordinate(vertexc)at(.5,.8)edge[]node[rightlabel,pos=.3]{a'}(vertexb) edge[blue,bend left]node[leftlabel,pos=1]{1}(.3,1.2) edge[bend right]node[rightlabel,pos=1]{a}(.7,1.2); \draw[red](-.1,0)to[out=-90,in=-90](-.5,0)to[wavyup](-.6,1.2)node[rightlabel]{1} (.3,0)to[out=-90,in=-90,looseness=1.5](-.7,0)to[wavyup](-.8,1.2)node[leftlabel]{c}; }.$$
By Proposition \ref{p:spinnetsignstrong}, the relation can be straightened, with the introduction of signs $S_1$ and $S_2$, to:
$$\label{eq:straightenstrands} S_1 \tikz[trivalent]{ \coordinate(vxa)at(.5,.4)edge[]node[rightlabel,pos=1]{b}(.5,-.1) edge[bend left]node[leftlabel,pos=1]{c}(0,1.1) edge[bend right]node[rightlabel,pos=1]{a}(1,1.1); \draw[blue](.2,1.1)to[out=-90,in=-90,looseness=2](.8,1.1)node[leftlabel]{1};} = \sum_{\substack{a'\in\iadm{a,1}\\ c'\in\iadm{c,1}\cap\iadm{b,a'}}} S_2\: \mfs_{a'}(a,1)\Fus ba{a'}c{c'} \tikz[trivalent]{ \coordinate(vxa)at(.5,.4)edge[]node[rightlabel,pos=1]{b}(.5,-.1); \coordinate(vxb)at(.1,.8)edge[bend right]node[leftlabel,pos=.5]{c'\:}(vxa) edge[bend left]node[leftlabel,pos=1]{c}(-.2,1.2) edge[blue,bend right]node[leftlabel,pos=1]{1}(.4,1.2); \coordinate(vxc)at(.9,.8)edge[bend left]node[rightlabel,pos=.5]{a'}(vxa) edge[blue,bend left]node[rightlabel,pos=1]{1}(.6,1.2) edge[bend right]node[rightlabel,pos=1]{a}(1.2,1.2);}.$$

To calculate $S_1S_2$, we must count the number of kinks in each diagram. No strands in \eqref{eq:straightenstrands} are kinked. The strands between $b$ and $c$ on the lefthand side of \eqref{eq:extendstrands} are kinked, giving a factor $S_1=\mfs_a(b,c)$. On the righthand side, strands between $b$ and either $1$ or $c$ are kinked, and so are those between $1$ and $c$, producing a factor $S_2=\mfs_{a'}(b,c')\mfs_{c'}(1,c)$. Hence
$$S_1S_2=\mfs_{c'}(1,c)\mfs_{a'}(b,c')\mfs_a(b,c)=(-1)^{\frac12(1-(a'-a))}=\mfs_{a'}(1,a).$$
Therefore, the coefficient is $\left(\mfs_{a'}(1,a)\right)^2\Fus ba{a'}c{c'}=\Fus ba{a'}c{c'}$.
\end{proof}
\end{proposition}

The reflective symmetries guaranteed by Proposition \ref{p:spinnetreflection} imply that corresponding formulas for the products
$$\label{eq:gluevertexlist} % \tikz[trivalent]{ % \coordinate(vxa)at(.5,.5) % edge[]node[rightlabel,pos=1]{a}(.5,-.1) % edge[bend left]node[rightlabel,pos=1]{c}(0,1.1) % edge[bend right]node[rightlabel,pos=1]{b}(1,1.1); % \draw[blue](.3,-.1)to[wavyup](-.2,1.1)node[leftlabel]{1};} % \quad \tikz[trivalent]{ \coordinate(vxa)at(.5,.5) edge[]node[leftlabel,pos=1]{a}(.5,-.1) edge[bend left]node[leftlabel,pos=1]{b}(0,1.1) edge[bend right]node[leftlabel,pos=1]{c}(1,1.1); \draw[blue](.7,-.1)to[wavyup](1.2,1.1)node[rightlabel]{1};} \quad \tikz[trivalent]{ \coordinate(vxa)at(.5,.5) edge[]node[rightlabel,pos=1]{a}(.5,1.1) edge[bend left]node[rightlabel,pos=1]{b}(1,-.1) edge[bend right]node[rightlabel,pos=1]{c}(0,-.1); \draw[blue](-.2,-.1)to[wavyup](.3,1.1)node[leftlabel]{1};} \quad \tikz[trivalent]{ \coordinate(vxa)at(.5,.5) edge[]node[leftlabel,pos=1]{a}(.5,1.1) edge[bend left]node[leftlabel,pos=1]{c}(1,-.1) edge[bend right]node[leftlabel,pos=1]{b}(0,-.1); \draw[blue](1.2,-.1)to[wavyup](.7,1.1)node[rightlabel]{1};}$$
can be generated by reflecting \eqref{eq:gluevertexleft}. In particular, the terms in these summations have the same coefficients.

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